Vector Mechanics for Engineers: Statics and Dynamics
Vector Mechanics for Engineers: Statics and Dynamics
12th Edition
ISBN: 9781259638091
Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self
Publisher: McGraw-Hill Education
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Chapter 12.3, Problem 12.104P

(a)

To determine

Find the increase in speed required at point A for the satellite to achieve the escape velocity and enter a parabolic orbit.

(a)

Expert Solution
Check Mark

Answer to Problem 12.104P

The increase in speed required at point A for the satellite to achieve the escape velocity and enter a parabolic orbit is 1.637×103m/s_.

Explanation of Solution

Given information:

The altitude of circular orbit of the satellite from the surface of the earth (r) is 19,110 km.

The radius of the earth (R) is 6,370 km.

Calculation:

Find the equation of product (GM) of the constant of gravitation G and the mass M of the earth using the equation:

GM=gR2

Substitute 9.81m/s2 for g and 6,370 km for R.

GM=9.81×(6,370km×1,000m1km)2=398.06×1012m3/s2

Find the altitude of circular orbit of the satellite (rA) from the center of the earth using the Equation:

rA=R+r

Substitute 6,370 km for R and 19,110 km for r.

rA=6,370+19,110=25,480km×1,000m1km=25.48×106m

Find the velocity of satellite (vcirc) in the circular orbit using the equation:

vcirc=GMrA

Substitute 398.06×1012m3/s2 for GM and 25.48×106m for rA.

vcirc=398.06×101225.48×106=3.9525×103m/s

Find the escape velocity of satellite (vcirc) from the circular orbit using the equation:

vesc=2GMrA

Substitute 398.06×1012m3/s2 for GM and 25.48×106m for rA.

vesc=2×398.06×101225.48×106=5.5897×103m/s

Find the decrease in speed (Δv) required at point A for the satellite to enter an elliptic orbit of minimum altitude 6370 km using the equation:

Δv=vescvcirc

Substitute 5.5897×103m/s for vesc and 3.9525×103m/s for vcirc.

Δv=(5.5897×103)(3.9525×103)=1.637×103m/s

Thus, the increase in speed required at point A for the satellite to achieve the escape velocity and enter a parabolic orbit is 1.637×103m/s_.

(b)

To determine

Find the decrease in speed required at point A for the satellite to enter an elliptic orbit of minimum altitude 6370 km.

(b)

Expert Solution
Check Mark

Answer to Problem 12.104P

The decrease in speed required at point A for the satellite to enter an elliptic orbit of minimum altitude 6,370 km is 725m/s_.

Explanation of Solution

Calculation:

Find the radius (rB) of the elliptical path.

rB=6,370+6,370=12,740km×1,000m1km=12.74×106m

Find the angular momentum per unit mass h using the equation.

1rA+1rB=2GMh2rA+rBrArB=2GMh2h2=2GMrArBrA+rBh=2GMrArBrA+rB

Substitute 398.06×1012m3/s2 for GM, 25.48×106m for rA, and 12.74×106m for rB.

h=2(398.06×1012)(25.48×106)(12.74×106)(25.48×106)+(12.74×106m)=82.230×109m2/s

Find the velocity at A (vA) in parabolic orbit using the equation:

h=rAvAvA=hrA

Substitute 82.230×109m2/s for h and 25.48×106m for rA.

vA=82.230×10925.48×106=3.2272×103m/s

Find the decrease (Δv) in speed required at point A for the satellite to enter an elliptic orbit of minimum altitude 6370 km using the equation:

Δv=vcircvA

Substitute 3.9525×103m/s for vcirc and 3.2272×103m/s for vA.

Δv=(3.9525×103)(3.2272×103)=725m/s

(c)

To determine

Find the eccentricity of the elliptic orbit.

(c)

Expert Solution
Check Mark

Answer to Problem 12.104P

The eccentricity of the elliptic orbit is 0.333_.

Explanation of Solution

Calculation:

Write the equation of angle at B.

θB=θA+180°

Apply cosine on both sides.

cosθB=cos(θA+180°)cosθB=cosθA

Find the constant C using the equation:

1rB1rA=CcosθBCcosθArArBrArB=C(cosθBcosθA)

Substitute cosθA for cosθB.

rArBrArB=C(cosθAcosθA)rArBrArB=2CcosθA

Substitute 180° for θA.

rArBrArB=2Ccos(180°)rArBrArB=2C(1)rArBrArB=2CC=rArB2rArB

Substitute 25.48×106m for rA and 12.74×106m for rB.

C=(25.48×106)(12.74×106)2(25.48×106)(12.74×106)=19.623×109m1

Find the eccentricity (ε) of the elliptic orbit using the equation:

ε=Ch2GM

Substitute 19.623×109m1 for C, 82.230×109m2/s for h, and 398.06×1012m3/s2 for GM.

ε=(19.623×109)(82.230×109)2398.06×1012=0.333

Thus, the eccentricity of the elliptic orbit is 0.333_.

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Chapter 12 Solutions

Vector Mechanics for Engineers: Statics and Dynamics

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